$\begin{array}{l} {a^{m} \cdot a^{n}=a^{m+n}} \\ \dfrac{a^{m}}{a^{n}}=a^{m-n} \\ a^{n}\cdot b^{n}=({ab})^{n} \\ (a^{n})^{m}=a^{n\cdot m} \\ \\ \textbf{Gegeben:} \\ {a=-\frac{1}{2} \qquad b=3 \qquad m=2 \qquad n=3}\\ \\ \textbf{Rechnung:} \\ {\left(-\frac{1}{2}\right)^{2} \cdot \left(-\frac{1}{2}\right)^{3}=\left(-\frac{1}{2}\right)^{2+3}=\left(-\frac{1}{2}\right)^{5}=-\frac{1}{32}}\\ \left(-\frac{1}{2}\right)^{2}:\left(-\frac{1}{2}\right)^{3}=\dfrac{\left(-\frac{1}{2}\right)^{2}}{\left(-\frac{1}{2}\right)^{3}}=\left(-\frac{1}{2}\right)^{2-3}=\left(-\frac{1}{2}\right)^{-1}=-2\\ \left(-\frac{1}{2}\right)^{3}\cdot 3^{3}=(\left(-\frac{1}{2}\right)\cdot3)^{3}= \left(-1\frac{1}{2}\right)^{3}={-3\frac{3}{8}} \\ (\left(-\frac{1}{2}\right)^{3})^{2}=\left(-\frac{1}{2}\right)^{3\cdot 2} = \left(-\frac{1}{2}\right)^{6}={\frac{1}{64}} \end{array}$